Question

Need help on these 2 an explanation of what formula’s to use would also be appreciated

Answer 1

17. .343 seconds: 18.a .357: 18.b 64.5N

Step-by-step explanation:

17. Since there is no friction, we can apply Newton's equation:

• 
  `f=ma`
• 
  `85N=5kg*a`
• 
  `17 m/s^2=a`

We can then use the kinematic equation:

• 
  `d=v_(i) *t+(1)/(2) at^2`
  `25N=U_(k) *70N`
• 
  `1m=0+(1)/(2)*17m/s^2*t^2`

Solve for t:

• 
  `{(1)/(8.5)s^2} =t^2`
• 
  `\sqrt{(1)/(8.5) }=t=.343 seconds`

18.a We can use the friction formula:

• 
  `F_(k) =U_(k) *F_(N)`
• 
  `25N=U_(k) *70N`

Solving for kinetic friction of block mass 7kg:

• 
  `.357=U_(k)`

18.b We should add all forces that will counteract the rightward acceleration.

• 
  `F_(a) =F_(x) -F_(N)`
• 
  `Fr_(7kg) =25N : Fr_(2.5kg) =15.75N`

Use Newton's equation:

• 
  `F=ma`

• 
  `F_(a) =9.5kg*2.5m/s^2`

• 
  `F_(a) =23.75N`

Solve for
`F_(x)`

`F_(a) +F_(fr) =F_(x)`

`23.75N+40.75N=F_(x) =64.5N`